00:01
Two different polynomial long division problems.
00:03
When we write our answers to each of these, we're going to write them in the form of a quotient and the remainder.
00:10
So i'm going to start with this first one.
00:12
We write our divisor on the outside, and we write the dividend or the thing being divided on the inside, just like we would with numerical long division.
00:28
The rest of this follows a very similar process to numerical long division as well.
00:32
Where we want to begin is we're going to take our 4x squared and divide by x.
00:38
So 4x squared divided by x is going to give us a 4x.
00:42
And the reason on putting this linear term on top of the other linear term is just for organizational purposes.
00:48
It helps me to kind of see when i am done and helps me stay organized throughout the problem.
00:54
So again, just like numerical long division, we now take this term and we'll multiply back by each of these here.
01:01
So 4x times x is a 4x squared, and 4x times negative 3 is a negative 12x.
01:08
And you'll see they line up nicely with like terms.
01:13
And we subtract 4x squared minus 4x squared cancels out.
01:17
3x minus negative 12x is a positive 15x.
01:22
And we drop down our next term.
01:24
And we're going to do it all over again.
01:27
This term is going to get divided by x.
01:30
So 15x divided by x is a positive 15.
01:35
Again, i have a constant being placed over a constant.
01:38
15 times x is 15x.
01:42
And 15 times negative 3 is a negative 45.
01:45
We put it in parentheses again and subtract.
01:48
15x minus 15x cancels out.
01:51
Negative 1 minus negative 45 is a positive 44.
01:57
So when we write our answer, we would write this as our, quotient 4x plus 15 plus our remainder 44 over our original divisor of x minus 3 and if you wanted to check your work to this what you could do is take your original divisor times your quotient and add your remainder and when you do that you'll find that you get the original polynomial that we divided let's take a look at the second one here we're going to be taking a quadratic and dividing by another quadratic.
02:45
So a little bit different from that last one, but overall the process is exactly the same.
02:52
I'm going to take our first term and divide it by the first term out here.
02:56
4x squared divided by x squared is just 4.
02:59
Since 4 is a constant, i'm going to lie that on top of the other constant.
03:03
4 times x squared is 4x squared, and 4 times negative 1.
03:09
I'm not going to put it under this linear term.
03:11
I'm going to put it underneath of the constant term back here.
03:14
We'll put it in parentheses and subtract.
03:17
4x squared minus 4x squared cancels out.
03:20
The negative 2x does not combine with anything.
03:23
And negative 23 minus negative 4 is a negative 19.
03:29
If we were to go further with this, we'd be trying to divide a negative 2x by x squared, which is not going to work out nicely for us.
03:39
So in addition to that, we kind of have that hint that we are done because this, there's no more room to write anything else up on the top.
03:47
So that divided out to be a quotient of 4 plus a remainder of negative 2x minus 19 over our original divisor.
03:57
So that looks like 4 times x squared minus 1 plus a negative 2x minus 19.
04:12
In this next problem, we're going to try synthetic division instead of long division.
04:17
We're going to take the quadratic 3x squared minus 2x plus 1 and divide it by x minus 1.
04:23
So when using synthetic division, we need to solve for what numbers should go in the box here.
04:33
So we set this factor equal to 0...